Question

Question: The vanadium in a sample of ore is converted to . The ion is subsequently titrated with in acidic solution to form and manganese (II) ion. – To titrate the solution, was required. If the mass percent of vanadium in the ore was what was the mass of the ore sample?

Short answer

Answer

The mass of the ore sample is $0.2613 \text{g}$

Step-by-step solution

Definition of ore sample

The process in which a portion (sample of ore) is selected in such a way that its composition will represent the average composition of the entire bulk of ore is defined as an ore sample.

Calculation of mass of the ore sample

The balanced equation is ${\text{5VO}}^{\text{2}}{\text{+ MnO}}^{\text{3}}{\text{+ 11H}}_{\text{2}}\text{O}\to {\text{Mn}}^{\text{2 +}}{\text{+ 5V(OH)}}_{\text{4}}^{\text{+}}{\text{+ 2H}}^{\text{+}}$

The mass of the

role="math" localid="1657713015602" $$\begin{gathered} {\text{V(MnO}}_{\text{4}}\text{)}=\text{26.45} \text{mL} \\ {\text{c(MnO}}_{\text{4}}\text{)} =0.0\text{2250} \text{M} \\ \text{c =}\frac{\text{n}}{\text{V}} \\ {\text{n(MnO}}_{\text{4}}\text{)}=\text{c.V} \\ ={\text{5.95.10}}^{-4} \text{mol} \end{gathered}$$

Molarity of :${\text{VO}}^{\text{2 +}}$

$$\begin{gathered} {\text{m(VO}}^{\text{2 +}}\text{)}={\text{n(VO}}^{\text{2 +}}{\text{).M(VO}}^{\text{2 +}}\text{)} \\ {\text{m(VO}}^{\text{2 +}}\text{)}=0.199 \text{g}\backslash \end{gathered}$$